Applied Drilling Engineering - Bourgoyne - CAP 4

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Chapter
Drilling Hydraulics
Chaps and provided information about the coinposi
tion and properties of drilling fluids and cement slurries
In this chapter the relation between the fluid properties
and the subsurface hydraulic forces present in the well
will be developed
The science of fluid mechanics is very important to the
drilling engineer Extremely large fluid pressures are
created in the long slender wellbore and tubular pipe
strings by the presence of drilling mud or cement The
presence of these subsurface pressures must be con
sidered in almost every well problem encountered In
this chapter the relations needed to determine the sub
surface fluid pressures will be developed for three com
mon well conditions These well conditions include
static condition in which both the well fluid and the cen
tral pipe string are at rest circulating operation in
which the fluids are being pumped down the central pipe
string and up the annulus and tripping operation in
which central pipe string is being moved up or down
through the fluid The second and third conditions listed
are complicated by the non-Newtonian behavior of drill
ing muds and cements Also included in this chapter are
the relations governing the transport of rock fragments
and immiscible formation fluids to the surface by the
drilling fluid
While it was not feasible to illustrate fully all the drill
ing applications of the fundamental concepts developed
in this chapter several of the more important applica
tions are presented in detail These applications include
calculation of subsurface hydrostatic pressures tend
ing to burst or collapse the
well tubulars or fracture ex
posed formations several aspects of blowout preven
tion displacement of cement slurries bit nozzle
size selection surge pressures
due to vertical pipe
movement and carrying capacity of drilling fluids
The order in which the applications are presented
parallels the development of the
fundamental fluid
mechanics concepts given in the chapter
This approach
best serves the needs of the beginning students ot drilling
engineering
4.1 Hydrostatic Pressure in Liquid
Columns
Subsurface well pressures are
determined most easily for
static well ponditions The variation of pressure with
depth in fluid column can be obtained by considering
the free-body diagram Fig 4.1 for the vertical forces
acting on an element of fluid at depth
in hole of
cross-sectional area The downward force on the fluid
element exerted by the fluid above is given by the
pressure times the
cross-sectional area of the element
F1 pA
Likewise there is an upward force on the element ex
erted by the fluid below given by
F2 p.A
In addition the weight of the fluid element is exerting
downward force gFven by
F3 FVAiD
where is the specific weight of the fluid Since the
fluid is at rest no shear forces exist and the
three forces
shown must be in equilibrium
pA
Expansion of the second term and division by the ele
ment volume AD gives
dpFdD 4.1
114
APPLIED DRILLING ENGINEERING
If we are dealing with liquid such as drilling mud or
salt water fluid compressibility is negligible and
specific weight can be considered constant with depth
Integration of Eq 4.1 for an incompressible liquid gives
pFDpo 4.2a
where the constant of integration is equal to the sur
face pressure D0 Normally the static surface
pressure Pc is zero unless the blowout preventer of the
well is closed and the well is trying to flow The specific
weight of the liquid in field units is given by
FO.O52p 4.3
where is the specific weight in pounds per square
inch per foot and is the fluid density in pounds mass
per gallon Thus Eq 4.2a in field units is given by
p0.052pDp0 4.2h
An important application of the hydrostatic pressure
equation is the determination of the proper drilling fluid
density The fluid column in the well must be of suffi
cient density to cause the pressure in the well opposite
each permeable stratum to be greater than the pore
pressure of the formation fluid in the permeable stratum
This problem is illustrated in the schematic drawing
shown in Fig 4.2 However the density of the fluid col
umn must not be sufficient to cause any of the formations
exposed to the drilling fluid to fracture fractured for
mation would allow some of the drilling fluid above the
fracture depth to leak rapidly from the well into the frac
tured formation
Example 4.1 Calculate the static mud density required
to prevent flow from permeable stratum at 12200 ft if
the pore pressure of the formation fluid is 8500 psig
Solution
Using Eq 4.2b
8500P-------------13.4 ibm/gal
0.052D 0.0a212200
Fig 4.2The well fluid system
Thus the mud density must be at least 13.4 lbm/gal to
prevent the
flow of formation fluid into the wellbore
when the well is open to the atmosphere psig
and there is no mud circulation
4.2 Hydrostatic Pressure in Gas Columns
In many drilling and completion operations gas
is
present in at least portion
of the well In some cases
gas is injected
in the well from the surface while in other
cases gas may enter the well from
subsurface fonna
tion The variation of pressure with depth in static gas
column is more complicated than in static liquid col
umn because the gas density changes with changing
pressure
The gas behavior can be described using
the real gas
equation defined by
pVznRTzRT 4.4
where
absolute pressure
gas volume
moles of gas
universal gas constant
absolute temperature
mass of gas
gas
molecular weight and
gas deviation factor
The gas deviation factor is measure of how much the
gas behavior deviates from that of an ideal gas An ideal
gas is one in which there are no attractive forces between
DO
Fig 4.1Forces acting on fluid element
ANN ULUS
DRILLING HYDRAULICS 115
gas molecules Gas deviation factors for natural gases
have been determined experimentally as function of
temperature and pressure and are readily available in the
petroleu00 literature
I3
In this chapter the simplifying
assumption of ideal gas behavior will generally be made
to assist the student in focusing more easily on the drill
ing hydraulics concepts being developed
The gas density can be expressed as function of
pressure by rearranging Eq 4.4 Solving this equation
for
gas density yields
pM
4.5a
zBT
where is expressed in pounds mass per gallon is in
pounds per square inch absolute and is in degrees
Rankine
When the length of the gas column is not great and the
gas pressure is above 1000 psia the hydrostatic equa
tion for incompressible liquids given by Eq 4.2b can be
used together with Eq 4.5b without much loss in ac
curacy However when the gas column is not short or
highly pressured the variation of gas density with depth
within the gas column should be taken into account Us
ing Eqs 4.1 4.3 and 4.5a we obtain
0.052 pM
dp dD
80.3 zT
If the variation in within the gas column is not too
great we can treat as constant Separating
variables in the above equation yields
P1
dD
P5
l544zT
D0
Integration of this equation gives
Example 4.2 well contains tubing filled with methane
gas molecular weight 16 to vertical depth of 10000
ft The annular space is filled with 9.0-lbm/gal brine
Assuming ideal gas behavior compute the amount by
which the exterior pressure on
the tubing exceeds the in
terior tubing pressure at 10000 ft if the surface tubing
pressure is 000 psia and the mean gas temperature is
140F If the collapse resistance of the tubing is 8330
psi will the tubing collapse due to the high external
pressure
4.3 Hydrostatic Pressure in
Complex Fluid Columns
During many drilling operations the well fluid column
contains several sections of different fluid densities The
variation of pressure
with depth in this type of complex
fluid column must be determined by separating the effect
of each fluid segment For example consider the com
plex liquid column shown in Fig 4.3 If the pressure at
the top of Section is known to be then the pressureat the bottom of Section can be computed from Eq
4.2b
Pm 0.052 p1D1 D0p0
The
pressure at
the bottom of Section is essentially
equal to the pressure at the top
of Section Even if an
interface is present the capillary pressure
would be
negligible for any
reasonable weilbore geometry Thus
the pressure at the
bottom of Section can be expressed
in terms of the pressure at
the top of Section
P2 0.052p2D2 --D10.052p 1D1 D0p0
In general the pressure
at any vertical distance depthD
can be expressed by
pp6O.052 pD1D1 4.7
Changing units from consistent units to common field
units gives
pM
p-
80.3zT
4.Sb
Solution The pressure in the annulus at depth
of
10000 ft is given by Eq 4.2b
P2 0.0529.0lO000-l- 14.74695 psia
The pressure in time tubing at depth of 10000 ft is given
by Eq 4.6
1610000
Pt l000e 154411460140 1188 psia
Thus the pressure difference is given by
P2Pt 469511883507 psi
which is considerably below the collapse pressure of the
tubing
The density of the gas in the tubing at the surface could
be approximated using Eq 4.Sb as follows
100016
0.331 Ibm/gal
80.31600
It is interesting to note that the use of this density in Eq
4.2b gives
0.0520.33 ll0000 10001172 psia
which is within 16 psi of the answer obtained using the
more complex Eq 4.6
MD D0
l544zT
Poe 4.6
116 APPLIED DRILLING ENGINEERING
Do
DLg
D4
It is frequently desirable to view the well fluid system
shown in Fig 4.3 as manometer when solving for the
pressure at given point in the well The drillstring in
terior usually is represented by the left side of the
manometer and the annulus usually is represented by the
right side of the manometer hydrostatic pressure
balance can then be written in terms of known pressure
and the unknown pressure using Eq 4.7
Example 4.3 An intermediate casing string is to be
cemented in place at depth of 10000 ft The well con
tains 10.5-Ibm/gal mud when the casing string is placed
on bottom The cementing operation is dcsigned so that
the 10.5-ibm/gal mud will be displaced from the annulus
by 300 ft of 8.5-Ibm/gal mud flush 1700 ft of
12.7-ibm/gal filler cement and 1000 ft of
16.7-ibm/gal high-strength cement The high-strength
cement will be displaced from the casing with 9-ibm/gal
brine Calculate the pump pressure required to complete
ly displace the cement from the casing
Solution The complex well fluid system is understood
more easily if viewed as manometer Fig 4.4 The
hydrostatic pressure balance is written by starting at the
known pressure and moving through the various fluid
sections to the point of the unknown pressure When
moving down through section D1 Di is positive
and the change in hydrostatic pressure is added to
the
known pressure conversely when moving up through
section D1 D1 is negative and the change in
hydrostatic pressure is
subtracted from the known
pressure
PaPO0M52
12.71 700 16.71 0009.0lO000
Since the known pressure Po is psig then
Pa266 psig
4.3.1 Equivalent Density Concept
Field experience in given area
often allows guidelines
to he developed for the maximum mud density that for
mations at given depth will
withstand without fractur
ing during normal drilling operations
It is sometimes
helpful to compare complex well fluid column to an
equivalent single-fluid column that is open to
the at
mosphere This is accomplished by calculating the
equivalent mud density Pe which is defined by
Pe
4.8
0.052D
The equivalent mud density always should be referenced
at specified depth
Example 4.4 Calculate the equivalent density at depth
of 10000 ft for Example 4.3 for static well conditions
after the cement has been displaced completely from the
casing
Solution At depth of 10000 ft
po
9pp
Brine
p0
atmospheric pressure
7000 feet
8.5 ppg
300 feet
1700 feet
I67p1 bOO feet.c
Fig 4.4---Viewing the well as manometer
on
Fig 4.3A complex liquid column
pO.0529OlO000 12665946 psig
DRILLING HYDRAULICS 117
Using Eq 4.8
5.946
Pe
_____________ 11.4 Ibm/gal
0.052 10000
4.3.2 Effect of Entrained Solids and Gases
in Drilling Fluid
Drilling engineers seldom deal with pure liquids or
gases For example both drilling fluids and cements are
primarily mixture of water and finely dividcd solids
The drilling mud in the annulus also contains the drilled
solids from the rock broken up by
the bit and the forma
tion fluids that were contained in the rock As long as the
foreign materials are suspended by the fluid or settling
through the fluid at their terminal velocity the effect of
the foreign materials on hydrostatic prcssurc can be corn
puted by replacing the fluid density in Eq 4.2b with the
density of the mixture However particles that have set
tled out of the fluid and arc supported by grain-to-grain
contact do not influence hydrostatic pressure
lhe density of an ideal mixture can be computed using
the calculation procedure discussed in Sec of Chap
The average density of mixture of several components
is given by
rn pV
i1 i1
i1
4.9
where andf are the mass volume density
and volume fraction of component respectively As
long as the components are liquids and solids the com
ponent density is essentially constant throughout the en
tire length of the column Thus the average density of
the mixture also will be essentially constant
If one component is finely divided gas the density of
the gas component does not remain constant but
decreases with the decreasing pressure drilling fluid
that is measured to have low density due to the
presence of gas bubbles is said to be gas cut
The determination of hydrostatic pressure at given
depth in gas cut mud can be made through use of the
real gas equation
If moles of gas are dispersed in or
associated with gal of drilling fluid the volume frac
tion of gas at given point
in the column is given by
NVRT
4.10
zNRi
In addition the gas density
at that point is defined by
Eq 4.5a Thus the effective density of the mixture is
given by
pfMNp
PPffgPgfg 4.11
zN1RT
where is the average molecular weight of the gas
For common field units substitution of this expression
for mean density in Eq 4.3 and combining with Eq 4.1
yields
D2 P2 pN.RT dp
0.052p1MN
4.12
If the variation of and is not too great over
the column
length of interest they can
be treated as con
stants of mean values and Integration of Eq 4.12
gives
D2D1P2I21 4.13
where
a0.052 pfMN 4.14
and
bNKr 4.15
It is unfortunate that the pressure P2 appears within the
logarithmic1 term in Eq 4.13 This means that an
iterative calculation procedure must
be used for the
determination of the change in pressure
with elevation
for gas-cut fluid column However if the gas/liquid
mixture is highly pressured and not very long the varia
tion of gas density with pressure can
be ignored In this
case the mixture density given by Eq 4.11 can be
assumed constant and the change in hydrostatic pressure
can be computed using Eq 4.2b
Exarnple 4.5 massive low-permeability sandstone
having porosity of 0.20 water saturation of 03 and
methane gas saturation of 0.7 is being drilled at rate
of 50 ft/hr with 9.875-in bit at depth of 12000 ft
14-lbm/gal drilling fluid
is being circulated at rate of
350 gal/mm while drilling Calculate the change in
hydrostatic pressure
caused by the drilled formation
material entering the
iiiud Assume that the mean mud
temperature
is 620R and that the formation water has
density of 9.0 lbm/gal
Also assume that the gas
behavior is ideal and that
both the gas and the rock cut
tings move at the same annular velocity as the mud The
density of the
drilled solids is 21.9 lbm/gal
Solution The hydrostatic head exerted by 12000 ft of
14-lbm/gal mud would be
pl4.70.OS2I4l2O00875l psia
The formation is being drilled at rate of
F9.87521 /7.48\
501 ------ 3.31 gal/mm
4144 60/
APPLIED DRILLING ENGINEERING
118
AVERAGE DENSITY OF GAS CUT MUD Ibm /aI
Using Eqs 4.14 and 4.15 gives
Il
a0.052 160.00023110.7312
2000
and
4000
4.7 79 blo.00023 620 11.5
As shown in the table belowvarious values forp2 were
6000
000
37
Since the well is open to the atmosphere the surface
pressure Pt is 14.7 psia The bottomhole pressure P2
8000 must be estimated from Eq 4.13 in an iterative manner
0000
39 1413
assumed until the calculated D2 was equal to the
well depth of 12000 ft
Fig 5Annular density plot for Example 4.5
P2
P2 P2P1
15.72 ln
psia 0.7312 Pi D2D1
Drilled solids are being added to the drilling fluid at
8750 11946 100.43 12046
rateof
8700 11878 100.34 11978
8716 11900 100.37 12000
3.3110.22.65 gal/mm
Thus the change in hydrostatic head due to the drilled
fonnation material entering the mud is given by
Formation water is being added to the drilling fluid at
rate of p871687Sl3S psi
3.310.20.30.2 gal/mm
The density of the drilling fluid after the addition of the
Example 4.5 indicates that the loss in hydrostatic head
water and drilled solids would be
due to normal contamination of the drilling fluid is usual
ly negligible In the past this was not understood by
143502l.92.6590.2 many drilling personnel The confusion was caused
mainly by severe lowering
of density of the drilling
3502.650.2
fluid leaving the well at the surface This lowering
of
density was due to the rapidly expanding entrained gas
14.057 ibm/gal
resulting from the decrease
in hydrostatic pressure on
the
drilling fluid as it approached
the surface The
Methane gas is being added to the drilling fluid at rate theoretical surface mud density that would be seen in Ex
of ample 4.5 is given by Eq 4.11 as
3.3l0.20.70.464 gal/mm
/5
Assuming the gas is ideal and the formation pressure is 14.7 l0.00023180.3620
approximately 8751 psia the gas density given by Eq
4.5b is
7.9 lbm/gal
875116
2.8 Ibm/gal As one driller remarked this amount of loss in mud den
80.31 .o0620
sity would cause even monkey to get excited In the
past it was common practice to increase the density of
Thus the gas mass rate entering the well is given by the drilling fluid when gas-cut mud was observed on the
surface because of fear of potential blowout
2.80.464 However Example 4.5 clearly
shows that this should
0.08 rnol/min not he done unless the well will flow with the pump off
16
As shown in Fig 4.5 significant decreases in annular
mud density occur only in the relatively shallow part of
Since the mud is being circulated at rate of 350 the annulus The rapid increase in annular density with
gal/mm the moles of gas per gallon of mud is given by depth occurs because the gas volume decreases by fac
tor of two when the hydrostatic pressure doubles For ex
0.081 ample increasing the hydrostatic pressure at the surface
N.O.OOO23l
mol/gal from 14.7 to 117.6 psia causes unit volume of gas to
350
decrease to one-eighth of its original size
DRILLING HYDRAULICS 119
4.4 Annuar Pressures During
Well Control Operations
One of the more important applications of the hydrostatic
pressure relationships is the determination of annular
pressures during well control operations Well control
operations refer to the emergency procedures followed
when formation fluids begin flowing into the well and
displacing the drilling fluid The flow of formation fluids
into the well is called kick schematic illustrating the
hydraulic flow paths during well-control operations is
shown in Fig 4.6 Formation fluids that have flowed in
to the weilbore generally must be removed by circulating
the well through an adjustable choke at the surface The
bottomhole pressure of the well at all times must remain
above the pore pressure of the formation to prevent addi
tional influx of formation fluid However com
plicating factor is the danger of fracturing weaker
stratum that also is exposed to the hydraulic pressure
Fracturing of an exposed stratum often results in an
underground blowout in which an uncontrolled flow of
formation fluids from the high-pressure stratum to the
fractured stratum occurs Thus the proper well control
strategy is to adjust surface choke so that the bot
tomhole pressure of the well is maintained just slightly
above the formation
pressure
plot of the surface annular pressure vs the volume
of drilling fluid circulated is called an annular pressure
profile Although annular pressure calculations are not
required for well control procedure used by most
operators today prior knowledge of kick behavior
helps in the preparation of appropriate contingency
plans Since annular frictional pressure losses are
generally small at the circulating rates used in well con
trol operations the calculations can be made using the
hydrostatic pressure equations
4.4.1 Kick Identification
The annular
pressure profile that will be observed during
well control operations depends to large extent on the
composition of the kick fluids In general gas kick
causes higher annular pressures
than liquid kick This
is true because gas kick has lower density than
liquid kick and must be allowed to expand as it is
pumped to the surface Both of these factors result in
lower hydrostatic pressure in the annulus Thus to main-
tam constant bottomiole pressure higher surface an
nular
pressure must be maintained using the adjustable
choke
Kick composition must be specified for annular
pressure calculations made for the purpose of well plan
ning Kick composition generally is not known during
actual well control operations However the density of
the kick fluid can he estimated from the observed
drillpipe pressure annular casing pressure
and pit gain
The density calculation often will determine if the kick is
predominantly gas or liquid
The density of the kick fluid is estimated most easily
by assuming that the kick fluid entered the annulus as
slug schematic of initial well conditions after closing
the blowout preventer on kick is shown in Fig 4.7
The volume of kick fluid present must be ascertained
from the volume of drilling fluid expelled from the an
nulus into the pit before closing the blowout preventer
The pit gain usually is recorded by pit volume
monitoring equipment If the kick volume is smaller than
the total capacity of the annulus opposite the drill collars
the length of the kick zone Lk can be expressed in
terms of the kick volume Vk and the annular capacity
C3 Assuming the well diameter is approximately con
stant we obtain
LkVkC3 4.16
where Lk is the length
of the kick zone V3 is the volume
of the kick zone and C3 is the annular capacity of the
hole opposite the
drill collars expressed as length per unit
volume
If the kick volume is larger than the total capacity of
the annulus opposite the drill collars then the length of
the kick zone Lk is given by
L3\
LkL3Vk__jC2 4.17C3
where L3 is the total length of the drill collars and C3 is
the annular capacity of the hole opposite the drill pipe
expressed as length per unit volume pressure balance
on the initial well system fora uniform mud density Pm
is given by
PUMP
Fig 4.6Schematic of well control operations Fig 4.7Scbonatjc of initial well conditions during well control
operations
Pc O.O52 PmTlPctp
120 APPLIED DRILLING ENGINEERING
blowout preventer
The volume of the kick-contaminated
zone can be estimated using
VGqt 4.20
thus allowing the mean density of the kickcontaminated
zone to be computed using Eq 4.18 The mean density
of the mixed zone then can be related to the density of
the kick fluid using the mixture equations Since
significant amount of natural mixing occurs even if the
pump is not operating when formation gas enters the
well Eq 4.20 tends to predict mixture volume that is
too low
Exaniple 4.6 well is being drilled at
vertical depth of
10000 ft while circulating 9.6-Ibm/gal mud at rate of
8.5 bbl/min when the well begins to flow Twenty bar
rels of mud are gained in the pit over 3-minute period
before the pump is stopped and the blowout preventers
are closed After the pressures stabilized an initial
drillpipe pressure of 520 psig and an initial casing
pressure of 720 psig are recorded The annular capacity
of thecasing opposite the drillpipe is 129 ft/bbl The an
nular capacity opposite the 900 ft of drill collars is 28.6
ft/bbl Compute the density of the kick fluid The total
capacity of the drillstring is 130 bbl
Solution schematic illustrating the geometry of this
example is given in Fig 4.8 The total capacity opposite
the 900 ft of drill collars is
900 ft
31.5 hbl
28.6 ft/bhl
720520
Pk96 2.9 Ibm/gal
0.052572
The results should be interpreted as an indication of low-
density kick fluidi.e gas
If it is assumed that the kick fluids are mixed with the
mud pumped while the well was flowing
Vk20 bbl8.5 bbl/min3 min45.5 hhl
The length of mixed zone is given by Eq 4.17 as
Lk90045531.Sl29lO8l ft
Using Eq 418 the mean density of the mixed zone is
given by
720 520
6.04 Ibm/gal
0.052t081
520
720
3500 ft
C0 2.9 ft/bbl
C0 2.9 ft/bbl
Dl0000 ft
P1 9.Sppg
Cdp 37 bbl
Mud-Gas Mixture
20 bbl of Gas
25.5 bbl of Mud
C0 28.6 ft/bbl
4.18 If it is assumed that the kick fluids entered as slug then
the volume of kick fluid is less than the total annular
capacity opposite the drill collars Thus
Fig 4.8Schematic for Example 4.6
Solving this expression for the density of the kick Pk
yields
Pc Pdp
Pk Pm _______
0.052
kick density less than about Ibm/gal should indicate
that the kick fluid is predominantly gas and kick densi
ty greater than about lbm/gal should indicate that the
kick fluid is predominantly liquid
Several factors can cause large errors in the calculation
of kick fluid density when the kick volume is small Hole
washout can make the determination of kick length dif
ficult In addition the pressure gauges often do not read
accurately at low pressures Also the effective annular
mud density may be slightly greater than the mud density
in the drillpipe because of entrained drilled solids Fur
thermore the kick fluid is mixed with significant quan
tity
of mud and often cannot be represented accurately as
slug Thus the kick density computed using Eq 418
should be viewed as only rough estimate
Some improvement in the accuracy of the kick density
calculation can be achieved if the volume of mud mixed
with the formation fluids is known The minimum mud
volume that was mixed with the kick fluids can be
estimated using the expression
Vqt 4.19
where is the flow rate of the pumps and td is the kick
detection time before stopping the pump and closing the
Lk 20 bbl 28.6 ft/bbl 572 ft
Using Eq 4.18 the density of the kick fluid is given by
DRILliNG HYDRAULICS 121
720
Pe 9.613.56 ibm/gal
0.0523500
After pumping 300 bbl of 10.6-ibm/gal mud the volume
of i0.6-ibm/gai mud in the annulus at the bottom of the
hole is
This is true since the total driuistring capacity is 130
bbl
The length of this region is given by
L1 90017o3l.5l2.92687 ft
4.4.2 Annular Pressure Prediction
The same hydrostatic pressure balance approach used to
identify the kick fluids also can he used to estimate the
pressure at any point in the annulus for various well con
ditions During well control operations the bottomhole
pressure will be maintained constant at value slightly
above the formation
pressure through the operation of an
adjustable choke Thus it is usually convenient to ex
press the pressure at the desired point in the annulus in
terms of the known hottomhole pressure This requires
knowledge of only the length and density of each fluid
region between the bottom of the hole and the point of
interest When gas kick is involved the length of the
gas region must be determined using the real gas equa
tion For simplicity it usually is assumed that the kick
region remains as continuous slug that does not slip
relative to the mud
The region above the new mud in the annulus will con
tain 130 bbl of 9.6-ibm/gal mud that was displaced from
the drillpipe The length of this region is given by
L213012.9 1677 ft
The region above the 9.6-ibm/gal mud will contain
methane gas The approximate pressure of the gas is
needed to compute the gas volume The pressure at the
bottom of the gas region can be computed from the
known bottomhole pressure which is 50 psi higher than
the formation pressure
Pg 5512 500.05210.62687
0.0529.6l 677 3244 psig
This pressure occurs at depth
of
10000268716775636 ft
Example Again consider the kick described in Ex
ample 4.6 Compute the mud density required to kill the
well and the equivalent density at the casing seat for the
shut-in well conditions Also compute the equivalent
density that would occur at the casing seat after pumping
300 hhl of kill mud while maintaining the bottomhole
pressure 50 psi higher than the formation pressure
through use of an adjustable surface choke Assume the
kick is methane gas at constant temperature of 140F
and that the
gas behaves as an ideal gas
Solution The formation pressure is given by
Phh 5200.0529.6l0000 5512 psig
The mud density required to overcome this formation
pressure is given by
5200.0529610000 520
______________ __________
0.05210000 0.05210000
pM 324414.716
Pg
80.3zT 8ft31460 140
1.08 ibm/gal
The region above the gas contains 9.6-ibm/gal mud
Thus we can compute the pressure at 3500 ft from the
known pressure of 3244 psig at 5636 ft using
p3244O.O52l .08437
Similarly the equivalent density
at the casing seat depth
of 3500 ft is given by
Since the column of mixed zone is only 1081 ft long and
under high pressure the mean density can be related to
the kick fluid density using equations for the effective
density of incompressible mixtures
2Op 25.59.6604
45.5
Solving this equation for the kick fluid density yields
Pk
20
1.5 ibm/gal
which also indicates that the kick fluid is gas
6.0445.59.625.5 300130170 bbl
For ideal gas
behavior the volume of the gas is given by
51214.7
20 bhl 33.9 hbl
3244 14.7
and the length of the gas region
is given by
Lg33.912943 ft
The density of the gas is given by Eq 4.5b
9.6 10.6 ibm/gal 0.0529.656364373500 2371 psig
122 APPLIED DRILLING ENGINEERING
4.5 Buoyancy
In the previous sections of this chapter we have focused
our attention on the calculation of hydrostatic pressure at
given points in the well In this section we will turn our
attention to the forces on the subsurface well equipment
and pipe strings that are due to hydrostatic pressure In
some cases only the resultant force or bending moment
created by hydrostatic pressure is needed while in others
the axial tension or compression at given point in the
submerged equipment is desired The determination of
the resultant force on submerged body will be con
sidered first
The net effect of hydraulic pressure acting on foreign
material immersed in the well fluid is called buoyancy
Buoyancy is understood most easily for vertical prism
such as the one shown in Fig 4.9 Hydraulic pressure
acting on side of the pnsm at any given depth is bal
anced by an equal hydraulic pressure acting on the op
posite side of the prism Thus the net force exerted by
the fluid on the prism is the resultant of force F1 acting
at the top and force F2 acting at the bottom of the prism
The downward force F1 is given by the hydraulic
pressure at depth times
the cross-sectional area
Fi p1AF.DA
Similarly the upward force on the prism is given by the
pressure at depth of Dh times the cross-sectional
area
F2 p2A FDhA
which indicates that the upward buoyant force Fh0 i.S
equal to the weight of the displaced fluid
This relation
was first used by Archimedes about 250 B.C
Archimedes relation is valid for foreign body im
mersed in fluid regardless of its shape For an
understanding of the more general ease of an irregular
body immersed in fluid
consider that the fluid
pressures at
the surface of the body would be unchanged
if the body were not present and this
surface were con
sidered an imaginary surface drawn in the liquid Since
the fluid element contained by the imaginary
surface is at
rest the sum of the vertical forces must
be zero and the
weight of the
contained fluid must be equal to
the
buoyantforce
The surrounding fluid acts with the same
system of forces on the foreign body
and this body also
experiences net upward force equal
to the weight of the
fluid region occupied by the bodyi.e equal to the
weight of the displaced fluid Note
that this same argu
ment could be applied even if the foreign body were im
mersed only partially in the fluid
The effective weight Wi of well equipment immersed
in fluid is defined by
WeWFbo 4.21
where is the weight of the well equipment in air and
Fh0 is the buoyant force
Using Archimedes relation the buoyant force is given
by
FbQpfVpf 4.22
PS
where
Pf
is the fluid density is the density of steel
and is the fluid volume displaced Substitution of Eq
4.22 into Eq 4.21 yields
WeWl_f2 4.23
The density of steel is approximately 490 lbm/cu ft or
65.5 Ibm/gal
Example 4.8 Ten thousand feet of 19.5-lbm/ft drillpipe
and 600 ft of 147-Ibm/ft drill collars are suspended off
bottom in 15-lbm/gal mud Calculate the effective
hook load that must be supported by the derrick
Solution The weight of the driuistring in air is given by
19.510000 147600 283200 Ibm
The effective weight of the drillstring in mud can be
computed using Eq 4.23
Thus the resultant buoyant force Fh0 exerted by the
fluid on the prism is given by
Fb0 F2 F1
We wl
15\
65.5
283200
218300
7__7_y7
F1
SURFACE OF LIQUID
F2
to PRISM Ib IRREGULAR BODY
Fig 49.-HydrauIic forces acting on foreign body
The equivalent density corresponding to this pressure is
given by
371
13.0 lbm/gal
0.0523500
FAD Is Fw5ADFe.vAh
DRILLING HYDRAULICS
123
Fig 4.10Effect of hydrostatic pressure on axial forces in drilistririg schematic of drillstririg free body diagramfor drill
collars and free body diagram for drillpipe
4.5.1 Determination of Axial Stress
In Example 4.8 only the net effect of hydrostatic
pressure on the pipe string was required However in
some cases it may be necessary to compute the axial
stress at given point in the pipe string The axial stress
is the axial tension in the pipe string divided by the cross-
sectional area of steel When axial stress must be deter
mined the cffcctive points of application of the
hydrostatic pressure must be considered and Ar
chimedes relation cannot be used
Consider the idealized schematic of drilistring
suspended in well Fig 4.lOa The lower portion of
the drilistring is composed of drill collars while the up
per portion is composed of drilipipe To apply
downward force Fb on the bit the drilistring is lowered
until portion of the weight of the pipe string is sup
ported by the bottom of the hole The cross-sectional
area of the drill collars A2 is much greater than the
cross-sectional area of the dnllpipe Note that
hydrostatic pressure is applied to the bottom of the drill
collars against cross-sectional area A2 and at the top
of the drill collars against the cross section A2 A1 To
determine the axial tension FT in the drill collars con
sider free body diagram of the lower portion of the
drillstring Fig lOb For the ystem to be in
equilibrium
FT_W2F2Fb_WdCXdCP2A2_Fb
where
4.24a
Wdc weight per unit length of drill collars in air
Xdc distance from the bottom of the drill collars
to the point of interest
P2 hydrostatic pressure at Point and
Fb force applied to the bit
To determine the axial tension in the drilipipe con
sider free-body diagram of the upper portion of the
drilistring Fig lOc As before system equilibrium
requires that
FTWI W2F1 F2Fb
op
FTwdxW2pIA2Al
P2A2Fb 4.24b
where
Wdp weight per
foot of drillpipe in air
Xdp
distance from the bottom of the drillpipe
top of drill collars to the point of in
terest and
Pt hydrostatic pressure at Point
jaltrss is obtained by dividing the axJjnslin by
the cross-sectiojiaLarea_oLsteel
Example 4.9 Prepare graph of axial tension vs depth
for the drillstring described in Example 4.8 Also
develop expressions for determining
axial stress in the
drilistring
Solution The hydrostatic pressure at the top of the drill
collars is given by
DRILLING MUD
DRILL PIPE
OAl
FT
dc
I-
O%2
F1
P1 O.05215lO0007800 psig
124
APPLIED DRILLING ENGINEERING
2000
4000
6000
800O
10000
12000
--COMPRESSION
3OOK
TENSION
HOOK LOAD
2I830O Ibf
DRILL PIPE
I95O00 Ibf
III
88200Ibf
F2 826843.2
357200 Ibf
Fig 4.11Axial tensions as function of depth for Example 4.9
Similarly the hydrostatic pressure P2 at the bottom of
the drill collars is given by
P2 0052 5l0600 8268 psig
The cross-sectional area of 19.5-ibm/ft driflpipe is given
approximately by
19.5 Ibm/ft
A1 144 sq in./sq ft5.73 sq in
490 Ibm/cu ft
Similarly the cross-sectional area of 147-ibm/ft drill
collars is given by
147 Ibm/ftA2-- 144 sq in./sq ft43.2 sq in
490 Ibm/cu ft
The tension in the drillpipe as function of depth is
given by Eq 4.24b
FT19.5lo000.D 147600
780043.2 5.73826843.2 0.0
After
simplifying this equation we obtain
FT218300_ 19.5D
for the 10000 ft range Note that this is the
equation of straight line with slope of 19.5 lbf/ft
and an intercept of 218300 lbf at surface Note that
this is the same result obtained in Example 4.8
At depths below 10000 ft Eq 4.24a can be used
FT 147l0600D826843.20
After simplifying this equation we obtain
FT 1201000 lbf147D
for the 10000 10600 ft range Using the equa
tions for FT as function of depth the graph shown in
Fig 4.11 was obtained
Axial stress in the pipe string is
obtained by dividing
the axial tension by the cross-sectional area
of steel
Thus axial stress az at any point in the drillpipe is given
by
21830019.5D
5.73
38098 psi3.403
foi the 0D 10000-ft range Similarly the axial
stress at any point
in the drill collars is given by
1201000147
43.2
27800 psi3.403
lOOK lOOK 300K
Fj 780 O4325.73
29213001bf
for the 10000D 10600-ft range
DRILLING HYDRAULICS 125
4.5.2 Effect of Buoyancy on Buckling
Long slender columns such as drilipipe have low
resistance to any applied bending moments and tend to
fail
by buckling when subjected to vertical compres
sional load As shown in Fig 4.12 if long slender
dnllpipe that is confined by welibore or casing is sub
jected to compressional load on bottom that is less than
the hook load helical buckling can occur in the lower
Portion of the pipe Buckling forces are resisted by the
moment of inertia of the pipe The moment of inertia of
circular
pipe is given by
IrI64d4 d4
ble Thus if buckling tendency exists above the drill
collars helical buckling may occur in the drillpipe as
shown in Fig 4.13
If drillpipe is rotated in buckled condition the
tool
joints will fatigue quickly and fail
It is common practice
to use enough heavy walled drill
collars in the lower sec
tion of the drillstring so that the desired weight may be
applied to the bit without creating tendency for the
drilipipe to buckle The point above which there is no
tendency to buckle is sometimes referred to as
the
neutral point At the neutral point the axial stress
is
equal to the
average
of the radial and tangential stresses
Fig 4.14 Current design practice is to maintain the
neutral point below the drillpipe during drilling
operations
If the
drilling fluid is air and the torque required
to
rotate the bit is low the radial and tangential stress in the
drillpipe may be negligible For these simplified condi
tions the neutral point is the point of zero axial stress
The length of drill collars then can be chosen such that
the weight of the collars is equal to the desired weight to
be applied to the bit In this case the minimum length of
drill collars Ldc is given by
Fb
Ld0-.-in air 4.25a
dc
where Fb is the maximum force to be applied to the bit
during drilling operations and Wdc is the weight per foot
of the drill collars Note that the use of this length of drill
collars under these simplified well conditions would
result in the neutral point occurring at the junction be
tween the drillpipe and drill collars No portion of the
slender drillpipe is subjected to axial compression
The effect of buoyancyon buckling should not be ig
nored if liquid drilling fluid is used However there
has been great amount of confusion in the past about
the proper procedure for including buoyancy into the
analysis Many people have reasoned that the net vertical
compressional force due to buoyancy simply should be
added to the compressional loading Fh when com
puting the minimum length of drill collars by Eq 4.25a
However this approach ignores the effect of hydrostatic
pressure on the radial and tangential stresses present at
the neutral point and thus may be overly conservative
For example this approach would predict the need for
considerable length of drill collars even for bit loading
Fb of zero
One of the easiest and most general approaches to in
cluding the effect of buoyancy on buckling was proposed
by Goins.4 Goins introduces stability force due to fluid
pressure p1
inside the pipe and pressure p0 outside the
pipe The stability force is defined by
F41pA0p0
where is the cross-sectional area computed using the
inside pipe diameter and A0 is the cross-sectional
area computed using the outside diameter of the pipe
The stability force can be plotted on tension/com
pression diagram such as the one shown in Fig 4.11
The neutral point then can be determined from the in- ./
tersection of the axial compression force and the stability
force
Neutral Point
Neutral Point
Fig 412Helical buckling of slender pipe in well slender pipe
suspended and partially buckled slender pipe
The preparation of graph of axial stress vs depth is
left as student exercise
where d0 is the nominal or outside diameter and is the
inside diameter For drill collars the moment of inertia
is large and generally is assumed to be great enough to
prevent
buckling However the moment of inertia of
dnlpipe is small and generally is assumed to be negligi
126
APPLIED DRILLING ENGINEERING
Ldc
Fig 4.13Helical buckling of drillpipe above drill collars
desired condition and undesired buckled condition
For simplified conditions when the fluid pressures are
due to the hydrostatic pressure of drilling
fluid of
uniform density corollary of Archimedes law can
be
applied Recall that
in developing
Archimedes relation
it was pointed out that the
fluid pressure acting on
the
surface of any foreign body
would be unchanged if the
body were not present and that
this surface was merely
an imaginary surface drawn in the liquid It was
reasoned
that if the fluid element contained by the surface is at
rest the sum of the vertical forces must
be zero and the
weight of the contained
fluid must be equal in magnitude
but opposite in direction to
the buoyant force However
it is also true that for the fluid element contained by the
surface to be at rest the sum of the moments acting on
the fluid element must be zero Thus the moment
caused
by the hydrostatic forces acting
on the fluid element must
be equal in magnitude
but opposite
in direction to the
moment caused by the weight of the contained fluid
regardless of the shape of the sutface
The weight of the
contained fluid in the imaginary surface
and the weight
of the foreign body both are distributed loads and
have
the same moment arm with respect to given point
This
means that for long slender column immersed
in fluid
the effective weight of the rod in
the fluid should be used
instead of the weight of
the rod in air when computing
bending moments Thus the proper length of drill col
lars Ldc required to eliminate tendency for
the
drillpipe to buckle is given by
Fb
Ldc in liquid 4.25b
Wdcl
It can be shown that the use of drill
collar length
predicted by Eq 4.25b for hydrostatic conditions will
result in an intersection of the stability force
line and the
axial compression line at the junction
between the
drillpipe and the drill
collars
Note that when Eq.4.25b is used only hydrostatic
pressures were
considered and the pressures due
to fluid
circulation are neglected Also neglected
are the effects
of the torque needed
for drillpipe rotation
These two
factors can have significant effect on the radial tangen
tial and axial stresses in the pipe wall and thus can
cause
significant shift in the
neutral point Also wall
friction
makes it difficult to determine the bit loading Fb from
the observed hook load Thus when Eq 4.25b is used
it is advisable to include safety factor of at least
1.3
Example 4.10 maximum bit weight of 68000 lbf is
anticipated in the next section of hole which starts at
depth of 10600 ft The drill collars have an internal
diameter of 3.0 in and an external diameter of 8.0 in
and the mud density is 15 ibm/gal Compute the
minimum length of drill collars required to prevent
buckling tendency from occurring in the 19.5-lbf/ft
drilipipe The drilipipe has an internal diameter of 4.206
in and an external diameter of 5.0 in Also show that the
intersection of plot of axial compressive stress
and
plot of the stability force occurs
at the top of
the drill
collars
Fb
Fh Fb
Neutral
Drill Pipe
Neutral Point
-Drill Collar
FbToo Large
Fig 4.14Stress state in steel
at neutral point
DRILLING HYDRAULICS 127
I-
Ui
COMPRESSION TENSION .- and the stability force at the surface is zero These points
-400 300K -200K -lOOK lOOK 200K 300K
have been used to plot stability force as
function of
in Fig 4.15 Note that the intersection of the axial
STABILITYJ
50300
2000- compression and stability force lines occur at the junc
STA8ILITY LOADING
The actual length of drill collars used in practice4000 FORCE
should be increased by applying safety factor Using
AXIAL tion between the drillpipe and drill collars
8000
safety factor of 1.3 gives
F2I83O0-6
F19.5D Ld6OOl.378O ft
8000
FTr 201 000 68 000L
00CC
4252OOL3D
200 4.6 Nonstatic Well Conditions
INSTGBILITY REGION
12000
The determination of pressure at various points in the
well can be quite complex when either the drilling mud
Fig 4.15Stability analysis plot for Example 4.10
or the drillstring is moving Frictional forces in the well
system can be difficult to describe mathematically
However in spite of the complexity of the system the
effect of these frictional forces must be determined for
Solution The weight per foot of the drill collars is given the calculation of the flowing bottomhole pressure or
by equivalent circulating density during drilling or cement
ing operations the bottomhole pressure or equivalent
7r82 32490
Wdc 147 lbflfl
circulating density during tripping operations the op
4144
timum pump pressure flow rate and bit nozzle sizes
during drilling operations the cuttings-carrying
The effective weight per foot in mud is given by capacity of the mud and the surface and downhole
pressures that will occur in the drilistring during well
15
113 lbf/ft
control operations for various mud flow rates
Wdce 147 The basic physical laws commonly applied to the
65.5
movement of fluids are conservation of mass
Thus the minimum length of drill collars required for
conservation of energy and conservation of momen
bit weight of 68000 lbf is given by
tum All of the equations describing fluid flow are ob
tained by application of these physical laws using an
68000 lbf
assumed rheological model and an equation of state Ex
Ld 600 ft ample rheological models used by drilling engineers are
113.3 lbf/ft the Newtonian model the Bingham plastic model and
the power-law model Example equations of state are the
Note that the well conditions are identical to those of Ex- incompressible fluid model the slightly compressible
ample 4.9 except that the load on the bit is 68000 lbf fluid model the ideal gas equation and the real gas
The stability force at the bottom of the collars is given by equation
F2Ap1A0p0 4.6.1 Mass Balance
The law of conservation of mass states that the net mass
rate into any volume is equal to the time rate of in-
328268 ____82 8268
crease of mass within the volume The drilling engineer
normally considers only steady-state
conditions in which357200 lbf the mass concentration or fluid density
at any point in the
well remains constant Also with the exception of air or
and the stability force at the top of the collars is given by gas
drilling the drilling fluid can be
considered incom
pressiblei.e the fluid density is essentially
the same
at all points in the well system In the absence
of any ac
7r
F2 .._....327800 _____827 800 cumulation or leakage of
well fluid in the surface equip-
ment or underground formations the flow rate of an in
compressible well fluid must be the same at all points in
337000 lbf the well
The mean velocity at given point is defined as the
Similarly the stability force at the bottom of the drillpipe
flow per unit area at
that point Because of nonuniform
is
flow geometry the mean velocity
at various points in the
well may be different even though
the flow rate at all
800____527800 points in
the well is the same knowledge of the mean
F5 _420627
velocity at given point
in the well often is desired For
example the drilling engineer frequently will compute
the mean upward flow velocity in the annulus to ensure44700 lbf
128 APPLIED DRILLING ENGINEERING
The energy leaving the system is the sum of
E2 p2 V2 enthalpy per unit mass of the fluid leav
ing the system at Point
gD2 potential energy per unit mass of the
fluid leaving the system at Point
kinetic energy per unit mass leaving the
system at Point
The work done by the fluid is equal to the energy per unit
mass of fluid given by the fluid to fluid engine or
equal to minus the work done by pump on the fluid
Thus the law of conservation of energy yields
E2 E1p2 V2 Pt V1gD2 D1
that it is adequate for rock-cutting removal Shown in
Table 4.1 are convenient forms of q/A for units frequent
ly used in the field
Example 4.11 12-Ibm/gal mud is being circulated at
400 gal/mm The 5.0-in drillpipe has an internal
diameter of 4.33 in and the drill collars have an internal
diameter of 2.5 in The bit has diameter of 9.875 in
Calculate the average velocity in the drillpipe
drill collars and annulus opposite the drilipipe
Solution Using the expression given in 1ab1e for
units of gallons per minute inches and feet per second
gives
400
Vdp 8.715 ft/s
2.4484.332
4.6.2 Energy Balance
The law of conservation of energy states that the net
energy rate out of system is equal to the time rate of
work done within the system Consider the generalized
flow
system shown in Fig 16 The energy entering the
system is the sum of
F1 p1 enthalpy per unit mass of the fluid
entering the system at Point
gD1 potential energy per unit mass of the
fluid entering the system at Point
kinetic
energy per unit mass of the fluid
entering the system at Point and
heat per unit mass of fluid entering the
system
P2 vWQ
Simplifying this expression using differential notations
yields
4.26
Eq 4.26 is the first law of thermodynamics applied to
steady flow process This equation is best
suited for flow
systems that involve either heat transfer or adiabatic
processes involving fluids whose thermodynamic prop
erties have been tabulated previously This form of the
equation seldom has been applied by drilling engineers
The change in internal energy of the fluid and the heat
gained by the fluid usually is considered using friction
loss term which can be defined in terms of Eq 4.26 us
ing the following expression
pdVQ 4.27
The frictional loss term can be used conveniently to ac
count for the lost work or energy wasted by the viscous
forces within the flowing fluid Substitution of Eq 4.27
into Eq 4.26 yields
VdpgDWF 428
Eq 4.28 often is called the mechanical energy balance
equation This equation was in use even before heat flow
was recognized as form of energy transfer by Carnot
and Joule and is completely general expression con
taining no limiting assumptions other than the exclusion
of phase boundaries and magnetic electrical and
chemical effects The effect of heat flow in the system is
included in the friction loss term
The first term in Eq 4.28
Pipe
17.16 bbl/min
d2
3.056 Cu fl/mm
2.448 d2
where
TABLE 4.1
Annulus
17.16 bbl/min
d-d
3.056 cu It/mm
dd
gal/rn in
2.448
and
average velocity ft/s
internal diameter of pipe in
d2 internal diameter of outer pipe or borehole
in and
external diamter of inner pipe in
400
Vdº 26.143 ft/s
2.4482.52
400
Vp 2.253 ft/s
2.4489.8752 52
DRILLING HYDRAULICS 129
Solution The average velocity in the drill collars is
400
26.14 ft/s
2.4482.52
Pb
8.074x 104p
DO
FL AL
may be difficult to evaluate if the fluid is compressible
unless the exact path of compression or expansion is
known Fortunately drilling engineers deal primarily
with essentially incompressible fluids having constant
specific volume
Since for incompressible fluids the term
Vdp
is given by
r2Vdp
Eq 4.28 also can be expressed by
p-pg ADppWpF
Expressing this equation in practical field units of
pounds per square inch pounds per gallons feet per sec
ond and feet gives
Pi 0.052pD2 D1--8.074
l04pf zipPfP2 4.29
Example 4.12 Determine the pressure at the bottom of
the drilistring if the frictional pressure loss in the drill-
string is 1400 psi the flow rate is 400 gals/mm the mud
density is 12 Ibm/gal and the well depth is 10000 ft
The internal diameter of the drill collars at the bottom of
the drillstring is 2.5 in and the pressure increase
developed by the pump is 3000 psi
ERGY rNrMG OUT WQD DONE
Fig 4.16Generalized flow system
4.7 Flow Through Jet Bits
schematic of
incompressible flow through
short con
stnction such as bit nozzle is shown in Fig 4.17 In
practice it generally is assumed that the change in
pressure due to change in elevation is negligible
the
velocity v0 upstream of the nozzle is negligible
compared with the nozzle velocity and the fric
tional pressure loss across the nozzle is negligible Thus
Eq 4.29 reduces to
Pt P2
Substituting the symbol Pa for the pressure drop
P1 P2 and solving this equation for the nozzle veloci
ty yields
4.30
Unfortunately the exit velocity predicted by Eq 4.30
for given pressure drop across
the bit never is
realized The actual velocity is always smaller than the
velocity computed using Eq 4.30 primarily because the
assumption of frictionless
flow is not strictly true To
compensate
for this difference correction factor or
discharge coefficient Cd usually is introduced so that the
modified equation
vnCdJb4 4.31
will result in the observed value for nozzle velocity The
discharge coefficient
has been determined experimental
ly for bit nozzles by
Eckel and Bielstein These authors
indicated that the discharge coefficient may be as high as
0.98 but recommended value of 0.95 as more prac
tical limit
ruck bit has more than one nozzle usually having
the same number of nozzles and cones When more than
one nozzle is present the pressure drop applied across all
of the nozzles must be the same Fig 18 According
to Eq 4.31 if the pressure drop is the same for each noz
zle the velocities through all nozzles are equal
The average velocity in the mud pits is essentially zero
P2 00.05212l00008.074
10 I226 142 30001400
062406.630001400
7833 psi
Example 4.12 illustrates the minor effect of the kinetic
energy term of Eq 4.29 in this drilling application In
general the change in kinetic energy caused by fluid ac
celeration can be ignored except for the flow of drilling
fluid
through the bit nozzles
130 APPLIED DRILLING ENGINEERING
Fig 4.17Flow through bit nozzle
Doll StrIng
Therefore if the nozzles are of different areas the flow
rate through each nozzle must adjust so that the ratio
q/A is the same for each nozzle If three nozzles are
present
qi q2 q3v_---_
A1 A2 A3
Note also that the total flow rate of the pump is given
by
qqt q2q3i5A1 -l-tA2i5A3
Simplifying this expression yields
qDA1 A2A3vA
Thus the velocity of flow through each nozzle is also
equal to the total flow rate divided by the total nozzle by
area
Fig 4.18Flow through parallel nozzlesExample 413 12.0-ibm/gal drilling fluid is flowing
through abit containing three %2-in nozzles at rate of
400 gal/mm Calculate the pressure drop across the bit
Solution The total area of the three nozzles is given by
A1 132 132132
4322
767 11r4169169169
0.3889 sq in
Using Eq 4.34 the pressure drop across the bit is given
4.32
Pb
A1 A1 A2 A1
In field units the nozzle velocity is given by
1169 psi
4.33
3.117 A1
where has units of feet per second has units of
gallons per minute and A1 has units of square inches
Combining Eqs 4.31 and 4.33 and solving for the
pressure drop across the bit zpb yields
8.311 X105pq2Ph
CA12
4.34
Since the viscous frictional effects are essentially
negligible for flow through short nozzles Eq 4.34 is
valid for both Newtonian and non-Newtonian liquids
Bit no7zle diameters often are expressed in 32nds of an
inch For example if the bit nozzles are described as
12-13-13 this denotes that the bit contains one nozzle
having diameter of in and two nozzles having
diameter of 32 jfl
4.7.1 Hydraulic Power
Since power is the rate of doing work pump energy
can be converted to hydraulic power PH by multiplying
by the mass flow rate pq Thus
PH pWqpq
If the flow rate is expressed in gallons per minute and
the pump pressure Ip is expressed in pounds per square
inch
4.35
1714
where is expressed in hydraulic horsepower
Likewise other terms in Eq 4.29 the pressure balance
equation can be expressed as hydraulic horsepower by
multiplying the pressure term by q/l714
____/ .--- --
_7_
Nozzle
8.311 XI05124002
0.9520.38892
DRILLING HYDRAULICS 131
Example 414 Determine the hydraulic horsepower be
ing developed by the pump discussed in Example 12
How much of this power is being lost due to the viscous
forces in the drillstring
Solution The pump power bcing used is given by Eq
4.35
Lpq 3000400
700 hp
1714 1714
The power consumed due to friction in the drilistring
is
Lpfq 1400400
_____ 327 hp
1714 1714
4.7.2 Hydraulic Impact Force
The
purpose of the jet nozzles is to improve the cleaning
action of the drilling fluid at the bottom of the hole
Before jet bits were introduced rock chips were not
removed efficiently and much of the bit life was con
sumed regrinding the rock fragments While the cleaning
action of thc jet is not well-understood several in
vestigators have concluded that the cleaning action is
maximized by maximizing the total hydraulic impact
force of the jetted fluid against the hole bottom If it is
assumed that the jet stream impacts the bottom of the
hole in the manner shown in Fig 4.17 all of the fluid
momentum is transferred to the hole bottom Since the
fluid is traveling at vertical velocity before striking
the hole bottom and is traveling at zero vertical velocity
after striking the hole bottom the time rate of change of
momentum in field units is given by
in pq
F1 4.36
32.1760
where pq is the mass rate of the fluid Combining Eqs
4.31 and 4.36 yields
F/0.01823CqJ 4.37
where
F1
is given in pounds
Example 15 Compute the impact force developed by
the bit discussed in Example 4.13
Solution Using Eq 4.37
F1 0.01823O954OO5lJ69
820 lbf
4.8 Rheological Models
The frictional pressure loss term in the pressure balance
equation given as Eq 4.29 is the most difficult to
evaluate However this term can be quite important
since
extremely large viscous
forces must be overcome
to move drilling fluid through the long slender conduits
used in the rotary drilling process
mathematical
description of the viscous forces present
in fluid is re
quired for the development of friction loss equations
The
rheological models generally used by drilling
engineers to approximate fluid behavior are the
Newtonian model the Bingham plastic model and
the power-law model
4.8.1 Newtonian Model
The viscous forces present in simple Newtonian fluid
are characterized by the fluid viscosity Examples of
Newtonian fluids are water gases and high gravity oils
To understand the nature of viscosity consider fluid
contained between two large parallel plates of area
which are separated by small distance Fig 19
The upper plate which is initially at rest is set in motion
in the direction at constant velocity After sufficient
time has passed for steady motion to be achieved con
stant force is required to keep the upper plate moving
at constant velocity The magnitude of the force was
found eperimentally to be given by
The term F/A is called the shear stress exerted on the
fluid Thus shear stress is defined by
4.38
Note that the area of the plate is the area in contact
with the fluid The velocity gradient v/L is an expression
of the shear rate
dv
4.39
dL
Thus the Newtonian model states that the shear stress
is directly proportional to the shear rate
4.40
where the constant of proportionality is known as the
viscosity of the fluid Fig 4.20 In terms of the moving
plate this means that if the force is doubled the plate
velocity also will double
Viscosity is expressed in poises poise is
dyne-s/cm2 or g/cms In the drilling industry
132 APPLIED DRILLING ENGINEERING
dinitially
at rest
//7\\./// \\//7\\7/\\77\\///\\///\\7
to
Small
Velocity buildup
in unsteady flow
Final velocity
distribution in
steady flow
Large
Fig 4.20.Shear stress vs shear rate for Newtonian fluid
viscosity generally is expressed in centipoises where
cp 0.01 poise Occasionally viscosity is expressed in
units of lbf-s/sq ft The units of viscosity can be related
at sea level by
lbf-s 454 g/lbf980 cm/s2
ft2 30.48 cm/ft2
479 dyne-s/cm2
479 poise 47900 cp
10 cm/s
10 seconds
cm
Using Eq 4.40
dyne/cm2
0.5 dyne s/cm2
10 seconds
or
The linear relation between shear stress and shear rate
described by Eq 4.40 is valid only as long as the fluid
moves in layers or laminae fluid that flows in this
manner is said to be in laminar flow This is true only at
relatively low rates of shear At high rates of shear the
flow pattern changes from laminar flow to turbulent
flow in which the fluid particles move downstream in
tumbling chaotic motion so that vortices and eddies are
fonned in the fluid Dye injected into the flow stream
thus would be dispersed quickly throughout the entire
cross section of the fluid The turbulent flow of fluids
has not been described mathematically Thus when tur
bulent flow occurs frictional pressure drops must be
determined by empirical correlations
Example 4.16 An upper plate of 20-cm2 area is spaced
cm above stationary plate Compute the viscosity in
centipoise of fluid between the plates if force of 100
dyne is required to move the upper plate at constant
velocity of 10 cm/s
Solution The shear stress is given by
100 dyneT5 dyne/cm2
20 cm2
4.8.2 Non-Newtonian Models
Most drilling fluids are too complex to be characterized
by single value for viscosity The apparent viscosity
measured depends on the shear rate at which the
measurement is made and the prior shear rate history of
the fluid Fluids that do nut exhibit direct propor
tionality between shear stress and shear rate are
classified as non-Newtonian Non-Newtonian fluids that
are shear-rate dependent Fig 4.21 are pseudoplastic if
the apparent viscosity decreases with increasing shear
Upper plate set
in motion
tO
Fig 4.l9Larninar flow of Newtonian fluids
The shear rate is given by
rn
tx
p.
tx
Ct
SHEAR RATE
50 ep
DRILLING HYDRAULICS 133
Lu
Cr
Cr
Lii
cr
Lu
V1
SHEAR RATE
Fin 4.21Shear stress vs shear rate for pseudoplastic and dilatant fluids pseudoplastic behavior /a2 and
dilaiant behavior P2 ai
Fig 4.22Shear stress vs time for thixotropic and rheopectic fluids thixotropic behavior and rheopectic behavior
rate and are dilatant if the apparent viscosity increases
with increasing shear rate Drilling fluids and cement
slurries are generally pseudoplastic in nature
Non-Newtonian fluids that are shear-time-dependent
Fig 4.22 are thixotropic if the apparent viscosity
decreases with time after the shear rate is increased to
new constant valueand are rheopectic if the apparent
viscosity increases with time after the shear rate is in
creased to new constant value Drilling fluids and ce
ment slurries are generally thixotropic
The Bingham plastic and power-law rheological
models are used to approximate the pseudoplastic
behavior of drilling fluids and cement slurries At pres
ent the thixotropic behavior of drilling fluids and ce
ment slurries is not modeled mathematically However
drilling fluids and cement slurries generally are stirred
before
measuring the apparent viscosities at various
shear rates so that steady-state conditions are obtained
Not accounting for thixotropy is satisfactory for most
cases but significant errors can result when large
number of direction changes and diameter changes are
present in the flow system
4.8.3 Bingham Plastic Model
The Bingham plastic model is defined by
ILp YTy 4.4la
rrr 4.4th
and
r/i 4.41c
graphical representation of this behavior is shown in
Fig 4.23
Bingham plastic will not flow until the applied shear
stress exceeds certain minimum value known as
the yield point After the yield point has been exceeded
changes in shear stress are proportional to changes in
shear rate and the constant of proportionality is called the
plastic viscosity /1 Eqs 4.41a through 4.4lc are valid
only for laminar flow Note that the units of plastic
viscosity are the same as the units of Newtonian or
parent viscosity To be consistent the units of the yield
Cl
C/
Lu
Cr
Cl
Cr
Lu
21
Lu
Cr
Cr
Ui
TIMET TIMET
134 APPLIED DRILLING ENGINEERING
Fiç 4.23Shear stress vs shear rate for Bingham plastic fluid
point must be the same as the units for shear stress
Thus the yield point has consistent units of dynes per
square centimeter However yield point usually is ex
pressed in field units of pounds per 100 sq ft The two
units can be related at sea level by
lbf 454 g/lbf980 cm/s2
100 sq ft 10030.48 cm/ft2
4.79 dyne/cm2
Example 4.17 An upper plate of 20-cm2 area is spaced
cm above stationary plate Compute the yield point
and plastic viscosity of fluid between the plates if
force of 200 dynes is required to cause any movement of
the upper plate and force of 400 dynes is required to
move the upper plate at constant velocity of 10 cm/s
or
lOOcp
4.8.4 Power-Law Model
The power-law model is defined by
rKj 4.42
graphical representation of the model is shown in Fig
4.24 Like the Bingham plastic model the power-law
model requires two parameters for fluid characterization
However the power-law model can be used to represent
pseudoplastic fluid Newtonian fluid
or dilatant fluid Eq 4.42 is valid only for
laminar flow
The parameter usually is called the consLctency in
dex of the fluid and the parameter usually is called
either the power-law exponent or the flow-behavior in
dex The deviation of the dimensionless flow-behavior
index from unity characterizes the degree to which the
fluid behavior is non-Newtonian The units of the con
sistency index depend on the value of Khas units of
dyne-s/cm2 or g/cms2 In this text unit called
the equivalent ccntipoisc eq cp will be used to represent
0.01 dyne-s7cm2 Occasionally the consistency index
is expressed in units of lbf-s1/sq ft The two units of
consistency index can be related at sea level by
lbf-s 454 g/lbf980 cm/s2
sq ft 30.48 cm/ft2
479 dyne-sYcm2
47900 eq cp
Solution The yield point is given by Eq 4.41a with
200 dyne 10 dyne/cm2
20 cm2
In field units
10T2.O9 lbf/l00 sq ft
4.79
The plastic viscosity is given by Eq 4.41a with
given by
10 cm/s10 seconds
cm
Thus is given by
400/20 10
10
1.0 dyne-s/cm2
Example 4.18 An upper plate of 20 cm2 is spaced cm
above stationary plate Compute the consistency index
and flow-behavior index if force of 50 dyne is required
to move the upper plate at constant velocity of cm/s
and force of 100 dyne is required to move the upper
plate at constant velocity of 10 cm/s
Solution Application of Eq 442 at the two rates of
shear observed yields
50 /4\fl
20
and
100 /10\
20
Dividing the second equation by the first gives
100
50
SHEAR STRESS
DRILUNG HYDRAULICS
135
Fig 4.24Shear stress vs shear rate for power-law fluel pseudoplastic power-law fluid and dilatant power-law fluid
Taking the log of both sides and solving for yields
log 100/50
0.756
log 10/4
Substituting this value of in the first equation above
yields
50 dyne-s756
0.8765
2040756 cm2
87.65
eq cp
4.9 Rotational Viscometer
Examples 4.16 through 4.18 illustrate the physical
meaning of the Newtonian Bingham plastic and power-
law parameters Unfortunately it would be extremely
difficult to build viscometer based on the relative
movement of two Oat parallel plates However as shown
in Fig 4.25 the rotation of an outer sleeve about con
centric cylinder is somewhat similar to the relative
movement of parallel plates The viscometer described
in the standard API diagnostic tests for drilling fluids
see Chap is rotational viscometer
Rotation of the outer sleeve instead of the inner bob
has been found to extend the transition from laminar to
turbulent flow to higher shear rates Since only the
laminar flow regime can be described analytically all
fluid characterization measurements must be made in
laminar flow In practice the torque exerted by the fluid
on the stationary bob usually is measured by torsion
spring attached to the bob The rotor and bob dimensions
available on the rotational viscometer are shown in Table
4.2 The r11/r22 rotor/bob combination is the stan
dard combination used for field testing of drilling fluid
The fluid shear rate between stationary and moving
parallel plates was assumed to be constant in Examples
4.16 through 4.18 However the fluid shear rate in
rotational viscometer is function of the radius The
fluid vclpcity at given radius is related to the angular
velocity by
vrw 4.43
Thus the change in velocity with radius is given by
dv dwrw
dr dr
If the fluid layers were not slipping past one another but
moving together as solid plug the change in velocity
with radius would be given by
dv
no slip
dr
Thus the shear rate due to slippage between fluid layers
is given by
dc
4.44
dr
When the rotor is rotating at constant angular veloci
ty and the bob is held motionless the
torque applied by the
torsion spring to the bob must be
equal but opposite
in direction to the torque applied to
the rotor by the motor The torque is transmitted between
the rotor and the bob by the viscous drag between suc
cessive layers of fluid If there is no slip at the rotor wall
the layer of fluid immediately adjacent to the rotor also is
moving at an angular velocity Successive layers of
fluid between r2 and r1 are moving at successively
lower velocities If there is no slip at the bob wall the
layer of fluid immediately adjacent to the bob is mo
tionless If the small end effect at the bottom of the bob
Lii
Lii
1-
SHEAR RATE SHEAR RATE
136 APPLiED
DRILLING ENGINEERING
TABLE 4.2ROTOR AND BOB DIMENSIONS
FOR ROTATIONAL VISCOMETERS
Rotor Sleeve
Bob
fluid
Bob Dimensions
Radius Length
Type cm cm
r11 1.7245 3.80
r12 1.2276 3.80
T3 0.86225 3.30
r1 0.86225 1.89
Rotor Dimensions
Radius
Type ciii
r21 1.8415
r22 1.7539
r33 2.5867
Assuming that no slip occurs at the walls of the
viscometer the angular velocity is zero at r1 and 032 at
r2 Thus separating variables in Eq 4.46 gives
360.50 dr
dw 4.47
27rh
r1
r3
Integrating and solving for viscosity yields
360.5071 1\
4.48a
4irhw2
Substituting the value 2ir N/60 for w2 the values of r1
r2 and shown in Table 4.2 and changing viscosity
units to centipoise simplifies this equation to the
following
300 4.48b
where
fluid viscosity cp
dial reading of the rotational viscometer and
speed of rotation of the outer cylinder rpm
Note that if the rotational viscometer is operated at 300
rpm the dial reading of the viscometer is numerically
equal to the viscosity in centipoise Most viscometers
used in the field will operate at either 300 or600 rpm
However in some cases multispeed viscometer that
will operate at 100 200 300 and 600 rpm is used
It is often desirable to know the shear rates present in
rotational viscometer for given speed of
rotation Eq
4.48a can be rearranged to give
360.50
Fig 4.25Bottom view of rotational viscometer
is ignored then the torque Tcan be related to the shear
stress in the fluid at any radius between the bob radius
r1 and the rotor radius r2 using the following equation
TT 27rrh
The spring constant of the torsion spring generally used
in testing drilling fluids is chosen such that
T360.5
where is the dial reading of the Fann measured in
degrees of angular displacement Equating the two ex
pressions for torque and solving for shear stress yields
360.5
4.45
2rhr2
Eq 4.45 indicates that the shear stress present in the
fluid varies inversely with the square of the radius This
relation is consequence of the geometry of the
viscometer and does not depend on the nature of the
fluid The shear rate can be related to shear stress using
the defining equation for the Newtonian Bingham
plastic or power-law fluid models discussed in the
previous sections
4.9.1 Newtonian Model
If the fluid can be described by the Newtonian fluid
model then the shear stress at any point in the fluid is
given by
dw
Ly
dr
Combining this equation with Eq 4.45 yields
dw 360.50
4.46
dr 27rhr3p
Substituting this expression with Eq 4.46 yields
47rN/60
dr r30.l37
2r
r1 r2
DRILLING HYDRAULICS 137
Thus the shear rate is given by
dw 5.066N
____________
dr r2
Example 19 Fann viscometer normally is operated
at 300 and 600 rpm in the standard API diagnostic test
Compute the shear rate that would occur at the bob
radius of 1.7245 cm for these two rotor speeds if the
viscometer contained Newtonian fluid
Solution Using Eq 4.49 we obtain
5.ObbN
1.703N
1.72452
At rotor speed of 300 rpm the shear rate at the bob is
given by
1.703300511 seconds
Similarly at rotor speed of 600 rpm the shear rate at
the bob is given by
1.703600 1022 seconds
4.9.2 Non-Newtonian Models
The rotational viscometer can also be used to determine
the flow parameters of the Bingham plastic and power-
law fluid models The equations needed for the calcula
tions of these flow parameters can be derived by follow
ing the same steps used in the derivation of the equations
for the Newtonian model These derivations are
presented in Appendix and the final equations are
summarized in Table 43 Since two flow parameters
must be calculated for both the Bingham plastic and
power-law models two readings must be made with
rotational viscometer at different rotor speeds Normally
the 300 and 600 rpm readings are used in the computa
tion However when it is desirable to characterize the
fluid at lower shear rates the flow parameters can he
computed using readings taken at lower rotor speeds
Example 420 rotational viscometer containing non-
Newtonian fluid gives dial reading of 12 at rotor
speed of 300 rpm and dial reading of 20 at rotor speed
of 600 rpm Compute the consistency index and flow-
behavior index of the power-law model for this fluid
Solution Using Table 4.3 the flow-behavior index and
consistency index are given by
3.32 log20/120.737
51012
5ll 61.8 eq cp
4.10 Laminar Flow in Pipes and Annuli
The drilling engineer deals primarily
with the flow of
drilling fluids and cements down the circular bore of the
drilistring and up
the circular annular space between the
drillstririg and the casing or open
hole If the pump rate is
low enough for the flow to be laminar the Newtonian
Bingham plastic or power-law model can be employed
to
develop the mathematical relation
between flow rate
and frictional
pressure drop In
this development these
simplifying assumptions are made the drillstring is
placd concentrically in the casing or open hole
the
dnllstring is not being rotated
sections of open
hole
are circular in shape and of known diameter the
drill
ing fluid is incompressible and
the flow is
isothermal
In
reality none of these assumptions are completely
valid and the resulting system of equations
will not
describe perfectly the laminar flow of drilling fluids in
the well In addition the student should keep in mind
that the Newtonian Bingham plastic and power-law
fluid rheological models do not take into account the
thixotropic nature of drilling mud and only approximate
the actual laminar flow fluid behavior Some research
has been conducted on the effect of pipe eccentricity6
pipe rotation and temperature and pressure
variations78
on flowing pressure gradients However the additional
computational complexity required to remove the
assumptions listed above seldom is justified in practice
Fluid flowing in pipe or concentric annulus does
not have uniform velocity If the flow pattern is
laminar the fluid velocity immediately adjacent to the
pipe walls will be zero and the fluid velocity in the
region most distant from the pipe walls will be max
imum Typical flow velocity profiles for laminar flow
pattern arc shown in Fig 4.26 As shown in this figure
concentric rings of fluid
laminae are telescoping down
the conduit at different velocities Pipe flow can be con
sidered as limiting case of annular flow in which the in
ner radius of the pipe r1 has value of zero
relation between radius shear stress and fric
tional pressure gradient dpf/dL
can be obtained from
consideration of Newtons law of motion for shell of
fluid at radius Shown in Fig 4.27 is free-body
diagram of shell
of fluid of length z.L and of thickness
L\r The sign convention used
in Fig 4.27 is such that
the direction of flow is from left to right and that the
velocity of flow is decreasing
with increasing radius
Thus the next shell of fluid enclosed by the fluid ele
ment of interest is moving faster than the fluid element of
interest Furthermore the next shell of fluid enclosing
the element of interest is moving slower than the element
of interest The force F1 applied by the fluid pressure at
Point is given by
F1 p2rrLSr
4.49
or
0.618 dynes0737/Cm2
133 APPLIED DRILLING ENGINEERING
TABLE 43SUMMARY OF EQUATIONS FOR ROTATIONAL VISCOMETER
300 5.066
Newtonian Model
7ON vN
Bingham Plastic Model 3U0
or
300 5.066 479r 3.174
0JN 0N2 ___N___L__ _i
Ty oo
or
N1
TYON _tp
OmaxCt rpm
PowerLaw Model n3.322 log
0300
or
log..i
510
7O.2094N
Shy
or
5100
1703
Likewise the force F2 applied by the fluid pressure at If the fluid element is moving at constant velocity the
Point is given by sum of the forces acting on the elements must equal zero
Summing forces we obtain
dpfdL
27rthrp_27rrLrp____LL
The negative sign for the dpf/dL tenri is required
dT
because the frictional pressure change is used to 2rrLr2Lrrrr0
represent p1 P2 rather than P2 Pi dr
The frictional force exerted by the adjacent shell of
fluid enclosed by the fluid element of interest is given by Expanding this equation dividing through by
2irrL1rzL and taking the limit as r0 yields
F3 r2irthL
dpf drr
Similarly the fnctional force exerted by the adjacent 4.50
shell of fluid that encloses the fluid element of interest is dL dr
given by
Since dp/dL is not function of Eq 4.50 can be in-
F4 Trr tegmted with respect to Separating variables yields
dT dpcrdrr\rdr
dr dL
4.51
where C1 is the constant of integration Note that for the
special case of pipe flow the constant C1 must be zero if
the shear stress is not to be infinite at Eq 4.51
which relates shear stress and frictional pressure gradient
at given radius is consequence of the geometry
of the
system and does not require the assumption of fluid
rheological model
The shear rate for the sign convention used in the
derivation is given by
-_ 4.52
dr
The shear rate can be related to shear stress using the
defining equation for the Newtonian Bingham plastic
or power..law fluid model
4.10.1 Newtonian Mocld
If the fluid can be described with the Newtonian fluid
model the shear stress at any